Platitud y Salidas Planas

8/3/2019 Platitud y Salidas Planas

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Flatness, tangent systems and flat outputsJaume FranchMay 25, 1999


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m

UNIVERSITAT POLITCNICA CATALUNYA

To Zael and Carlota


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/ * V UWVER1TAT POLITCNICA"' J ! CAT'-UMYA

V

A c k n o w l e d g e m e n t s 1During the years leading up to this thesis, I have had the good fortune to meet many people towhom I wish to express my gratitude.Firs t of all, I would like to acknowledge my advisor, Enric Fossas. One day, believing th at Icould tackle a thesis, he introduced me to the world of control theory. He has maintained thisbelief throughout the years, even in the worse moments, when he encouraged me not to give up.Furthermore, he made it possible for me to attend conferences, where I was able to learn a lot.His continuing support, our infinite discussions on some topics, his patient revision of my work,have all been crucial in the realization of this thesis.I would like to thank all the members of the Department of Applied Mathematics and Telematicswho have helped me a t one time or anoth er, especially those of the research gro up in differentialgeometry, dyn amical sy stems and applications. I would like to single out P rofessor M iguelCarlos Muoz, from whom I learned a great deal concerning differential geometry, especially inthe subject he was lecturing on at that time at the Faculty of Mathematics and Statistics. I amalso grateful to Xavier Gracia, who occasionally helped me with DMpjXproblems.I must not forget Antoni Palau, a student who in his final year made some Maple programsrelated to Chapter 4 under my supervision. I would also like to thank Jeff Palmer, my Englishteacher for four years (and I hope for many more), for his revision and correction of the englishmanuscr ip t .I would also like to mention my parents. They provided me with a fine education, giving me theopportunity of studying mathematics, as well as ceaseless encouragement in my undergraduatestudies and the preparation of this thesis.Last but not least, I wish to express my infinite gratitude to Zael, my wife. Without her supportin good and bad moments, her patience when arriving home late because of work, or when sherealized that often when she was talking to me I was thinking about a problem that had arisedthat day, and her unfailing love, this thesis would have not been possible.

' Pa r t i a l ly suppor ted by CICYT under Gran t TAP94-0552-C03-02 and TAP97-0969-C03-01


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C o n t e n t s1 I n t r o d u c t i o n 1

1.1 Linear system s 11.2 Sta tic feedback linea rizatio n 21.3 D yna mi c feedback lineariz ation 21.4 Fla tne ss in th e differential algebraic sett ing 31.5 Co ntents and contrib ution s 4

2 L i n e a r s y s t e m s 72.1 Linear contro l systems 72.2 Exa mp les 122.3 Con trollability and observability 142.4 M odes, poles and zeros 162.5 Exa mp les 182.6 Fin al rem arks 22

2.6.1 Th e ma tchin g condition in sliding control mo de 222.6.2 Th e linear system interconnections 233 L inear izat ion o f nonl inea r s y s t e m s and flatness 29

3.1 Different typ es of lineariz ation s 293.2 Fla tne ss an d differential algeb ra 313.3 Linea rization by prolong ations 34

4 L i n e a r i z a t i o n u s i n g d i f f e r e n t ia l algebra 3 74.1 Introduc tion 374.2 Single-input system s 374.3 Stati c feedback linearization of mu lti-input systems 414.4 Dy nam ic feedback linearization for mu lti-input systems 454.5 Software 474.6 Exam ples 474.7 An analogous proc edur e using field extensions 52

vi i


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viii5 Linearization by prolongations of 2-input sys tems 55

5.1 Prolongations of m inputs are not necessary 555.2 Main result 595.3 Examples 66

6 Improvement of the bounds for 3-input sys tems 756.1 Main Result 756.2 Where do the bounds 2r + 1 and In 2 + r come from? 90

7 Linearization by prolongations of m-input sys tems 937.1 Main results 937.2 About the bounds 102

8 Conclusions and sugges tions for further research 1058.1 The differential algebraic approach 1058.2 Linearization by prolongations and possible extensions 106

A Introduction to differential algebra 115A.l Basics on differential algebra 115A.2 The Kahler differential 117

B Software package for Chapter 4 119


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Lis t o f F igu r e s2.1 Elem entary RLC -circuit worked in example 3 202.2 Mechanical system with springs and dam pers corresponding to examp le 4 . . . . 212.3 Not stabilizable example 232.4 Revers e block diag ram of 2.3 242.5 Ill-posed syste m 242.6 Parallel interconnection 252.7 Series interco nnec tion 252.8 Feedback interconnection 262.9 Exa mp le illustrating possible changes of rank 27

ix


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LIST OF FIGURES


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C h a p t e r 1I n t r o d u c t i o nFeedback linearization of nonlinear control systems is a problem on which several scientistshave been working during the last twenty years. Its importance lies in the fact that it enablesus to transfer the properties of a linear system to a nonlinear one, as well as to propose asimple solution to one of the main problems of automtica, which is stabilization around a giventrajectory of the system.Feedback linea rization is in general an open problem . The re are solutions for specific cases,such as linearization by static feedback, the equivalence between static feedback and dynamicfeedback linearization for single input systems, systems with m inputs and m + 1 state variables,some cases of systems without drift,For many years differential geometric tools have been used in order to solve the problem offeedback linearization . Notions such as Lie brackets a nd involutive fields or dis tribu tion s are t hemost common tools in this context. But to solve partial differential equations is also needed.In the nineties, a new way of tackling the problem was proposed. Th is me tho d, re lated tothe n otion of flatness, was introd uced in the differential algebraic setting . Th is settin g led tonew concepts, and it has implied the introduction of new concepts for linear and nonlinearsystem s. It allows us to deal w ith a greater nu mb er of problems th an th e classic framew ork.Some years after the first works on the differential algebraic setting were done, two new versionsof flatness appeared, one using the differential geometry of infinite jets, and the other in theexterior differential systems.

1.1 Lin ear sys tem sIn 1963, K aim an [36] introduc ed a new meth od for describing linear control system s. In th is firstwork the foundations for a good comprehension and a revision of the results known at that timecan be found. Keywords introduced by Kaim an are state variables, controllability, observab ility,realization, minimal realization . . . .At th e end of the sixties and the beginning of the seventies, the algebraic theory of linear co ntrolsystem s in an arb itrar y field was developed . Roucheleau [55] and Roucheleau, W ym an and

1


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2 CHAPTER 1. INTRODUCTION

Kaiman [56] studied the realization problem over commutative rings. The situation in the noncommutative case, treated in [60] by Sontag in 1976, is different because the Cayley-Hamiltonfails to hold.W illem's pap ers [64] - [67] mu st be m entioned h ere, because the geom etric concept of trajecto riesintroduced therein plays a crucial role, and allows us to deal with many questions withoutdist inguishing between inputs, outpu ts, sta tes and other variables. Also impo rtant was th ework of Br uno vsk y [5], who gave a classification of linear controllable system s. Since then , t heresearchers in this area have referred to the Brunovsky canonical form, both those who work inlinear systems and those who try to linearize nonlinear control systems.At the beginning of the nineties, Michel Fliess suggested a new algebraic treatment for linearcontrol systems ([10], [11], [15], [16], [20]). T he corne rston e of his work resides in the fact th atit enables us to put linear control theory in an algebraic setting which utilizes module theory ina more general manner than that commonly employed since Kaiman. According to Fliess, thesepapers sketch an attempt to rehabilitate Kalman's point of view in the new context of moduletheory. His work is based on a state variable representation, where the dynamics is strictly inthe Kaiman form, but where the output map not only involves the state but also the controlvariables and their derivatives. This is the frame we wish to use in Chapter 2.

1.2 S ta t ic feedback l ine ariz at ionSince 1973, the problem of linearization of continuous nonlinear control systems has been extensively studied. Krener [41] found conditions for linearizing a system by means of state spacediffeomorphisms. A particular type of sta te feedback transform ation w as first introdu ced byBrockett [4]. This was later generalized for single input systems by Su [61], who also related hisresults to the notion of relative degree. The problem for multi input systems was finally solvedby Hunt, Su and Meyer [31] and Jakubczyk and Respondek [34]. Their works used mathematical tools such as Lie brackets and involutive distributions. In fact, they proved the equivalencebetween the static feedback linearization and the rank and involutivity of certain distributions.The Kronecker indices [51] were also a fundamental to this procedure.For non static feedback linearizable systems, some authors have considered partial linearizations([42],[44]), as well as approximate feedbacks ([29], [32]).

1.3 D yn am ic feedback l inear iza t ionPa rtial feedback linearization is related to input- ou tpu t decoupling. Necessary and sufficientconditions are available for this problem. For linear systems it is known that those conditionscan be weakened if one allows for a dynam ic com pensator. Th is motivated the intr od uct ionof a nonlinear dynamic state feedback transformation, which is a generalization of the staticsta te feedback trans form ation. In [6] and [7], the problem of dynam ic feedback lineariza tionwas studied by C harlet, Levine and Marino . App roaching the problem from the differentialgeometric point of view, they showed that single input systems that are dynamically feedback


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1.4. FLATNESS IN THE DIFFERENTIAL ALGEBRAIC SETTING 3

linearizable are also statically feedback linearizable, and two very special cases of dynamicallyfeedback linearizable multi input systems are also given in [6]. In [7] they presented fairly generalsufficient conditions for a system to be dynamic feedback linearizable by prolongations, as wellas a necessary conditions. Unfortunately, as they also showed with examples, neither are thesufficient conditions necessary nor are the necessary conditions sufficient.Aranda-Bricaire, Moog and Pomet gave a different approach in [1] and [2]. They characterizedthe flat or linearizing outputs in their framework, the so-called infinitesimal Brunovsky form.Again, although their result establishes a sufficient condition for the existence of a dynamicfeedback transformation that linearizes the system, this condition is not necessary in general.Sluis and Tilbury [59] gave an upper bound on the number of integrators needed to linearize acontrol system, but they proved only the sharpness of the bound for systems with two inputs.Their work was based on exterior differential systems. In the same framework, Rathinam andSluis [53] obta ined a test for dynam ic feedback linearization by reductio n to single inp ut syste ms .

1.4 F l a t ne ss in th e d i f ferentia l a lg ebra ic se t t in gDifferential alge bra was estab lishe d by R itt [54], Ka plan sky [39] an d Kolchin [40]. W ha t i nte res tsus most about this theory is the differential field extensions. Fliess was the first to introducedifferential algebra into control theory for linear and nonlinear systems. One of the chief featuresof the utilization of differential algebra is the avoidance of explicit equations. This enables usto deal with a greater number of problems.Using the differential version of the theorem of the primitive element, Fliess proposed a generalized canonical form in [10]. This was followed by a series of papers as a result of his jointwork w ith L evine, M art in a nd Ro uchon . See, for ins tanc e, [13], [14], [17], [18], [19], [22], [45].In these papers some concepts such as flatness and defect were introduced. One major propertyof flatness is the existence of what the authors called flat or linearizing outputs. The system isfiat, if and only if, without integrating any differential equation, the state and input variablescan be directly expressed in terms of the flat outputs and a finite number of their derivatives.Flatness is best defined by not distinguishing between input, state, output and other variables.This stan dpo int m atches W illems' approach in [64] well. He did not m ake distinctions am ongthe different types of variables.Flatness might be seen as another nonlinear extension of Kalman's controllability. In fact, anyflat non linear sys tem is contro llable. In add ition , for linear system s, flatness is equivalen t tocontrollab ility. A set of flat ou tp ut s is the non linear analo gue of a basis of a free mo du le. Itmust be emphasized that from trajectories of the flat outputs, trajectories for the states and theinputs are immediately deduced.The relationship between the nonlinear theory (using differential field extensions) and the lineartheory, wh ich utilizes m odu les, is given by w hat is called the Ka hle r differential [35]. Th ismathematical tool is used in this context to compute the associated tangent system to a nonlinearone. This tangent system is linear. Therefore, one strategy to obtain the flat outputs could beto compute the tangent system, and to find out an integrable basis of this tangent system.


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4 CHAPTER 1. INTRODUCTION1.5 C on ten t s an d con t r ibu t ion sThe aim of Chapter 2 is to present the state of the art on linear control systems within theframework of module algebraic theory. Fliess' papers are collected, although some proofs andexamples are new. Among new proofs, we would like to emphasize the proof of proposition 1,which gives the equivalence between a linear control system in state-space representation andmodules over a ring of differential operators. The proofs of section 2.4 are extensions of knownproofs, including all the details required to make such proofs more clear. This chapter has beensubmitted as a survey to the journal Linear algebra and its applications.In Chapter 3 some background necessary for understanding the main results of this work isgiven. The different types of linearization are presented: namely, static feedback linearization,linearization by prolon gation s, dynam ic feedback linearization, and flatness. Som e know n resu ltsin this field are stated, with appropriate references to locate the proofs.Chapter 4 deals with the problem of fla tness in a nonlinear mult i input (m inputs) system. Inthe framework of differential algebra, the tangent system is used in order to find out the m thfia t output when m 1 flat ou tp uts have been guessed. T he quotien t of mod ules is crucial inthis procedure. The contributions in this Chapter include:

1. A new proof of the well known fact that linearization by static and dynamic feedback areequivalent for single-input systems.2. A new algorithm to linearize single-input systems, as well as an algorithm to linearize

multi- input systems by stat ic feedback.3. A theoretic procedure to linearize any multi-input systems, together with a software pack

age to carry ou t the com puta tion s. Once the system is linearized, a condition to checkwhether or not the system can be linearized via prolongations is also derived.

4. An ap plicatio n of the procedu re to a vertical take off and landing (V TO L) aircraft. Tw onew flat outputs have been obtained, and it is proven that these flat outputs can be foundjust by using prolongations.

Thes e results have been pu blished in two conferences. In 1997 SAA EI [25], which refers t o sta ticfeedback linearizatio n, and in 1998 AC C [26], which is related to dyn amic feedback linearizat ion. Some parts of this Chapter, together with a part of Chapter 5, have been submitted toAutomtica.In Chapter 5, the problem of linearization by prolongations of systems with two inputs is studied.A bound on the number of integrators needed to linearize a control system is obtained, using themost elementary tools of differential geometry, such as Lie brackets and involutive distributions.An algorithm derived from this result is applied to some examples, some of which were thoughtunti l now to be not l inearizable by prolongations. For instance, the V TO L and the plana r d uctedfan. A part of this Chapter will appear in 1999 SAAEI [28].Chapter 7 generalizes the results of Chapter 5 to an arbitrary number of inputs, improving theexistent bounds in the literature when the number of inputs is greater than or equal to four. It


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1.5. CONTENTS AND CONTRIBUTIONS 5also contains a new proof of the fact that, when linearization by prolongations are considered,not al l the inputs must be prolonged. This Chapter has been submitted to Systems and ControlLetters.In the case of three inputs, better results are given in Chapter 6. These results have appearedin 1999 IFAC [27].This work ends with the conclusions and some suggestions for future research.


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6 CHAPTER J. INTRODUCTION

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C h a p t e r 2L i n e a r s y s t e m sTh is Ch apt er is organized as follows: Section 2.1 is devoted to com paring two definitions oflinear control systems in order to show their equivalence, and examples are given at the end.In Section 2.3, controllability and observability are presented in the module formalism. Modes,poles an d zeros are trea ted in section 2.4. Some exam ples clarify the work. Finally, someapplications to sliding control and linear systems interconnections are explained.

2 .1 L i ne a r c on t ro l s y s t e m sThis section deals with two definitions of linear control systems, the classical one in state-variables and a new one using left modules ([10]). The equivalence between both definitions isshown and some examples are given.D e f i n i t i o n 1 A linear control system in state-space representation is a system described by:

X = A{t)X + B (t)UY = C(t)X

where U = {u\,..., um) E Rm, A(t) M nxn, B{t) M mX n, X = (x\,..., xn) S Rn an dY = (yi,...,yp)eRpX are called state variables, U are input variables and Y are output variables.D e f i n i t i o n 2 A linear control system using left mod ules is a left finitely-generated K[-^] -modu leA. ( K is supposed to be a field 1 )D e f i n i t i o n 3 A linear dynamic with input U (u\, . . . ,u n ) is a linear control system A (that isto say a left finitely-generated K[-^]-module) which contains and such that A / isa torsion module. An output Y = (yi , . . . ,y p) is a finite set of elemen ts of the system.

XK R or C for constant l inear control systems, otherwise K is a field of mer om orph ic function s

7

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8 CHAPTER 2. LINEAR SYSTEMSP r o p o s i t i o n 1 The ab ove definitions are equivalent in the sense that if a system a s in definition1 is given, the left module of the definition 2 can be built, fulfilling the desired conditions.Conve rsely, if a system as in definition 2 is given, a realization as in 1 can be obtained.Proof. Fi rs t we see th at definition 1 implies definition 2.Let X = A(t)X + B{t)U (2.1)be a linear system as in definition 1. Consider the left r[g]-module generated by X and U.That is to say

M = K[^-]On the other hand, let N be the submodule generated by the relations of 2.1. Consider

A = M/Nthe quot ient submodule . As M is finitely generated ( a finite number of X and U ), A will alsobe finitely generated. So, it remains to be shown that A/ is torsion. Let z G M. T h e n ,

z = aiXi + ... + anxn + &iiii + ...bmumwhere ai,bi G K[^). Consider the natural projection over A,

z = a\X\ + ... + anxn + b\Ui + ...bmumBy con struc tion, any element of A has this form. M aking the quo tient A / we get:

z = a\X\ + ... + anxnIf the torsio n elem ents m ake up a su b mo dule, it is only necessary to show th at x~ l is torsion Vi.In order to end the proof the following lemmas are stated (the proof will be performed later):L e m m a 1 Vx 3P G K{^\ such that

PiXi G t f [ 4 ] atThis lemma sta tes tha t Xi is torsion in A/ .L e m m a 2 The torsion elements make up a submodule.Now, it m ust b e proven tha t definition 3 implies definition 1. As will be seen in lemm a 4,A TOP, where T is a torsion submodule and F is a free submodule.Let be {x}"1 a set of generators of F and {ZJ}^ a set of generators of T. Zj are torsionelements, so there exists Qj(j) such that

Qi(4)* i=0


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2.1. LINEAR CONTROL SYSTEMS 9On the other hand, there exists a submodule U such that A/7 is torsion. That is to say:

V z 3 P i ( | ) | Piij^xi E UAn output y is an element of A. So

For all , j let the next integers be defined by:di = max{degree(Pi),degree(Ri) + 1}ej max{degree(Qj),degree(Sj) + 1}

Then, for any i we have a system of the form: 1 2%i 3-j

di1

fc=0whe re a^ = a; , th e coeffiecients a come from th e eq ua tion Pj (^ )x G U or the(d degree(Pi)) th derivative of this equation, if necessary, and u = -F(g) o r a l s 0 * n e (* ~~degree{Pi))th if necessary.The same can be done for zy .

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10 CHAPTER 2. LINEAR SYSTEMSThat is to say, the dynamic generated by the realization obtained is the dynamic A/U.

Now we are going to prove the two lemmas previously stated.Proof of lemma 1: First case: A is a matrix with constant coeficients. Then, PA ( the charac-teristical polynomial of A ) accomplishes

p 4 ? x ' * K Indeed, in A/ the system is only

X = AXThe solution of this system is X(t) = eAtXo. If PA is applied to this system we obtain

PA(^)X = PA(jt)e AtXo = PA(A)eAtXQ = 0Where the Cayley-Hamilton theorem has been applied to the last equality. So, each componentof X, labelled x t fulfills PA{)X = -Since X(t) = eAtXo in the non constants coefficients case cannot be assured, this demonstrationdoes not hold. The corresponding equation to x must be derived, replacing the other variablesby their corresponding equations. If this method is i terated un ti l the n th derivative of X { , weobtain:

/ z i \ / 0 \ / l \ / 0 \Xi

\ * S n ) JHi x \ - \ H

V < WXi -\ h

v < / W W+ Xn

This is a (n+l)-vector which is a linear combination of n vectors. For this reason these vectorsare linearly dep end ent. Hence their determin ant is vanishing and this determin ant yields apolynom ial with inde term inate ^ such tha t when it is applied to x, it is zero.Obviously this can be done for any state-variable. Thus we can state that

Vi 3P i | PiXi = 0in A/ . In other words, x is torsion in the quotient subm odu le. Example: Consider the system

' t i \ . ( o sX = 1 t l X + \ 1 U

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2.1. LINEAR CONTROL SYSTEMS 11Let A be < Xi,X2,U >

< X\ tX\ X2, X2 X\ tX2 U >Following the above algorithm, torsion of x\ in A/ will be prov en. x\ fulfills the followingequations ( the same notat ion is used for x\ in A or in A/U):

X\ = X\X\ tX\ + X2

Deriving this equation: X\ = x\ + tx\ + 2. And m aking the sub st i tut ion for 1 and 2, i tbecomes

1 = (2 + 2)xi + 2tx 2 (2.2)The fol lowing system can be wri t ten

So the following determinant is vanishing:Xl1

01

xi t2 + 2 ItThat is

1 - 2ti + ( i 2 - 2)2:1 = 0in A/U. So there exists a polynom ial with indermin ate ^ tha t voids xi in A/U. The same canbe done for 2.Proof of lemma 2: The torsion elements must make up a submodule. Let T be the set of torsionelements. Two condit ions must be proven:

\/x,y eT=>x + y &T (1)VxeT,keK[~]^kxeT (2)at

First of all a property of the ring of differential operators A = K[-^\ will be proven:L e m m a 3

V a , e A,a^0,b^0,3a',b' \0b'a = a'b


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12 CHAPTER 2. LINEAR SYSTEMSProof: Let be

t0 j = 0If there exist,

n j m 7"' = 'The equali ty o'6 = b'a must be verified; that is to say:

(E4(|)')w|)')-(Ei(5),)a:-(5) i)fc=0 j=0 =0 t=0

Equaling term to te rm, a system with n+m + 1 homogeneous equations and n + m+2 unknownsis obtained. So it has a non trivial solution, and hence the existence of a' and b' fulfilling therequired conditions can be deduced. HOnce this fact has been proven, it is not difficult to prove lemma 2: As x G T, 3a G A \ ax = 0Analogously, as y G T, 3b G A \ by = 0. Using the property just proved 3a', b' | a'b = b'a ^ 0.Then b'a(x + y) = b'ax + b'ay = b'ay = a'by = 0. S o i + y e T .On the other hand, a'kx = k'ax = 0 = kx G T. This fact finishes the proof of the lemma 2. Now, anoth er lem ma will be sta ted and proven. This lemma will be useful in order to decomposethe submodule A into a direct sum of a torsion submodule and a free submodule.L e m m a 4 A = T F, where T is a torsion submodule and F is a free submodule ( that is tosay, without torsion elements ) .Proof: Consider the canonical morphism:

n : A A/rBy definition this morphism is linear and exhaustive. It is clear that the kernel of th is morphismis T. So it must must ensured tha t A/T is free. Let y ^ 0 be a torsion element of A/T.Such element is the image by the morphism II of an element y G A. As y is a torsion element3pi G K[\ | piy = 0 T h u s p\y G T and, consequently, 3p2 S K[-^] | P2P\V = 0. This is thesame as y is torsion or, in other words, y = 0, which contradicts our init ial assumptions. Thus,there is no torsion element in A/T; t ha t is to say A/T is a free module. F will be generated bythe one element of each subset of inverse images of the generators of the free module A/T.

2 . 2 E xa m pl e sl.

i = X22 = U

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EXAMPLES 13In th is example , M = < Xi,X2,u >. N is generated by the above equations. So,

< Xi,X2,U >A = =N = < Xx >When the quotient A/ is done, it can be seen that $pX\ = 0. Then an elementof the ring of differential operators that cancels X\ is obta ined . So A / is torsionbecause its only generator is torsion.

i = i i + u2 = 3 3 = u

In this example, M =< xi,X2,x,u >. N is also generated by the above equations. So< Xi,X2,X,U > < 1,2:2 >

N < 1 x\ +u,X2 3,2:3 u > In A/ the relat ions (^ I)x\ = 0 and ^ 2 :2 = 0 are sat isfied. Th us, the quotientmodule is torsion again.

/ 0

A =

\

10 1

0

X = AX + BU

\

0 10 1

00 1

0 10 )

B =

(0 0 \0 01 00 00 00 10 00 0

\o 0)Notice that ^2:3 = u\, $pX2 = u\ and ^3-2:1 = u\. The same happens to 2:4, 2:5 ,XQ andU2- Fu rth erm or e, 2:7, 2:8, X9 are torsion elem ents. There fore, it can be w ritt en

< Xi,...,Xg,Ui,U2 >< 1 X2, 2 2:3, 3 Ui, ... >

< X\, Xi, X7 > _ < X7 >' - = < 2:1,2:4 > e< 4 3 ) > < 4 3 ) >So the sy stem has a torsion su bmo dule, generated by 2:7 and a free sub mo dule gen eratedby 2:1 an d 2:4. N otw iths tan din g, A / < u\,U2 > is a torsion submodule.

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14 CHAPTER 2. LINEAR SYSTEMS4.

5.

Th is is an ex am ple of a no n-co nsta nt coefficient. T he ring over which is defined th e sy stemis R{t)[-ji], where R(t) is the field of fractions of real polyno mials. Th e system is rewri ttenin the following way:. _ < Xi,X2,U >

The relat ion

xi = ( - tl)xican be deduced from the first equation. And from the second equation

u = (jt - tl)x2 - Ix x = ((j t)2 - 2tjt + ( 2 - 2)I)X1So it can be said that A = < z i >So , It is clear that, using the latest relations written, A/ is a torsion module.

x = x + u (2 )The derivatives of the input are not considered in the state-space representation. Let bey = x u . y is a generator of the module and the module is free. The equation in thevariable y is :

y = y + u

2.3 Co nt ro l lab i l i ty an d observabi l i tyIn this section, simple characterizations of controllability and observability based on moduletheory techniques are given. See [10] again. The equivalence between control systems in state-space form and control systems in module theory will be used.T h e o r e m 1 A system A is controllable if and only if it is a free mo dule.Proof. First assu me th at A is free. If the system is uncontrollable, as in [36], we have a K aim anrealization as follows:

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2.3. CONTROLLABILITY AND OBSERVABILITY 15where Ai is a square matr ix and belongs to the uncontrollable part . Now lem ma l can be applied.Therefore, the elements of X\ are torsion. So there exists a contradict ion with the freeness ofA.Now assume that the system is controllable. Let A = F T be a descomposition in a directsum of a free left module and a torsion left module. If A was not free, T ^ 0. Let be

T =T h u s Vx 3Pj G K[^[ \ P{Xi = 0, because X{ are torsion elements. Th us a system with kequations of order nfc (polynomial degree) is obtained. This can be transformed into a systemwith order 1 equations where the state variables are xi, ...,xitJll, ...x^^, ...,Xk,nk This is anexpression of the form:

* i = AXXXHere there is a further contradict ion because this is an uncontrollable K aima n realization. Thecontradiction comes from the assumpt ion of non-freeness of the module. Th e above proof also shows an equivalence betw een the torsion subm odule and the uncontrollablepa r t of the Kaiman realization.I t can also be seen that , if the system is controllable, each element of A is related, directly orindirectly, to the inputs: directly, if it can be expressed as a linear combination of the inputs ;or indirectly, because each state-variable accomplishes a differential equation where there areinputs .Next the relation between observability and module theory is shown. In the classical theory, asystem:

X = AX + BUY = CX

is called observable if and only ifrank < C\ {A1 + ^-)C\..., {A1 + ^-^C* >= nat at

where n is the dimension of the state-variable vector.T h e o r e m 2 A system is observable if and only if

A=That is to say, if and only if each variable ofA can be written as a linear combination of inputs,outputs and their derivatives.Proof. Let Y = (yi,...,yp). The re is no loss of generality in assuming that j / i , . . . , yp are linearlyindependent. First of all, we suppose that A =< U, Y > or, in other words, < U, Y >=< U,X >.This is also equivalent to _

~

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16 CHAP TER 2. LINEAR SYSTEMSSo, the Kaiman realization is written in the quotient, where the variables are overlined:

~X = AX

Y = CXDeriving k t imes:

Consider the linear system obtained by gathering the former equation for A; = 0 , . . . , n 1. Sincewe have assumed that j / i , . . . ,y p are linearly independent, the system has a unique solution if,and only if,

rank{C\ {A1 + ~)C\..., {A* + ^T^C 1) = nat atwhich is the classical observability condition. Therefore,

rank(C\ (A' + jC\..., (A4 + | ) n - x C * ) = nif, and only if, X are written as a unique linear combination of Y and their derivatives; if, andonly if, X are written as a unique linear combination of U , Y and their derivatives.Notice that , from this proof, an equivalence between the observability part of the Kaimanrealization and the submodule follows.

2 .4 M o de s , po l es and ze rosIn this section we attempt to give an algebraic interpretat ion of the hidden modes, poles andzeros of the co nsta nt linear system s. It follows [11], althoug h some proofs have been e xtend ed.Let us recall the Kaim an realizations in the uncon trollable and in the unob servable cases. Inthe uncontrol lable case the Kaiman realization is:

U ) - ( 5 . ) ( s H ^ >where A\ is the uncontrollable part matrix. And in the unobservable case:

()-( SOUKS)*- ( f ) ( 5 )

where A3 is the unobservable part matrix.

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2.4. MODES, POLES AND ZEROS 17Classically, the hidden modes were the eigenvalues of A\ (input-decoupling zeros ) and A$(output-decoupling zeros). W hat is the interpretat ion of these ma trix in the module theory?Let us begin with A\:As this is the uncontrollable part matrix, A is not free. So the module can be decomposed in adirect sum of its free par t and its torsion p art: A = F T. Let us denote the linear mappinginduced by th e derivative ^ by

T:T>TTh is m app ing is well defined bec ause ^ is an element of the ring over which the mo dule A hasbeen defined, and T is a submodule of A. Recall also the equivalence between th e uncon trollablepart and the torsion submodule T. So A\ is the matrix of r, and therefore the input-decouplingzeros are the eigenvalues of the mapping T.Analogously, note that there is an equivalence between the observable part and < U, Y >.Consider the quotient submodule S = A / < U, Y >. Obviously there is an equivalence betweenthis quotient submodule and the unobservable part . Denote the l inear mapping induced by Jby: a:S>5This mapping is again well-defined and its matrix will be A3. Therefore the output-decouplingzeros are the eigenvalues of this mapping.Now, an interpretation for poles is looked for. Let be

A = A / r =/ 

which is torsion, because A/ is also torsion. As in the above interpretations, denote theAdt : A

l inear m app ing induced by 4 by:

Again remember that poles are the eigenvalues of A3 ( the controllable par t m atrix ) . Th en,since a; G A , t he equation

x = A3Xis satisfied, because when the quotient is done, the U part vanishes. Thus we obtain the proofthat poles are the eigenvalues of .Lastly, let us look for an interp retatio n of zeros. Consider the greatest torsion subm odu le in< U,Y > and call it Ti. Let be a quotient module. Notice that < U,Y > / < T\ > is free because the torsion part hasbeen removed. If when the quotient by < Y > is done, a torsion submodule appears, then the

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18 CHAPTER 2. LINEAR SYSTEMSquotient submodule J will be a dynamic where Y are now the inputs. More precisely, thereexist two polynomial matrices P{-j) and Q(^) , in such a way tha t

Now, it is straightforward procedure to obtain a realization of the system such asZ = Z + BYU = CZ

This is the inverse system, where U are the outputs and Y the inputs. We know that zeros arethe poles of th e inverse system. On the other hand den ote the linear mapping induced by $ by:e : J J

The poles of the inverse system are the eigenvalues of this mapping and, therefore, the zeros ofthe initial system are the eigenvalues of e.2 . 5 E xa m pl e s

1. Consider the linear control system described by

XI - 4 - 4 0 - 1 - 2 \1 0 0 0 00 0 - 4 - 5 - 20 0 1 0 0V o o o i o y

x +( o i \0 0- 1 00 0V o o /

u

Y 1 0 0 0 0- 1 0 1 0 0 X

= < X2,X5 >The module description of the system is

_ < Xi,X2,X3,X4,X5,U\,U2 >< equations >This is a free module and the system is therefore controllable. On the other hand it is cleartha t A/ is torsion. The observability of the system can also be checked: x\ = j / i ,*3 = /1 + V2, Z4 = (a* + 4 + 4 / )yi + (i + V - &uz> ^ 5 = - 1 / 2 $ ( i / i + y 2) - 4(yi +2/2) - 5x 4 i and x- i = - l / 4 ( $ a ; i + 4 a ; 1 + 4 + X5 + ^2)- In short < X,U >< Y, U >.So, the system is observable.Now, the derivatives of u\ and ui are written as functions of the outputs:

2 ,d? Ad2 d nTdT^ = -{d^ + Aa + 5dl+2I)^+y2)

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2.5. EXAMPLES 19d2 fd3 d2 Ad. ,d ,

The submodule J such tha t J =< U, Y > / < U, Ti > is, in this example, the same asJ =< U,Y > I < Y > . By the above equalities it can be affirmed that J is a torsionmo dule. And the m atrix of the mapping e is

A =( 0 1 0 0 \

0 0 0 00 0 0 1

V 0 0 0 0 JT hu s, it is clear tha t the re is only one eigenvalue, which is zero. Mo reover, it can b echecked that this is the unique zero of the system. See [38] for more details.

2. Consider the linear control system described by an aircraft altitude dynamics ( [58] ).

X = I - 40V 6

1 0 0 \- 4 0 00 0 10 0 0 )

x + ( \30W /

Now the module of the system isy = ( 0 0 1 O )

< equations >In order to find a generator for this module, the next procedure is shown:Wri te z = axi+bx2+cx2+dx4 :. Th en, the following condition must be imposed: derivativesup to order three cannot contain the control variable u. With this condition a generator ofthe system is z = 15xi + 2^2 + +2IZ3 + 6x4, because the following equalities are verified:

, (2 )Xl = 98

98142 4 i -2 2

*(2)X3 =

4

294Uz - AzW - z&

294So, z is a gen erator of A. Moreover, A = < z > is a free module or, in other words, thesystem is controllable.

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20 CHAPTER 2. LINEAR SYSTEMS

Figure 2.1: Elementary RLC-circuit worked in example 3Let us recall that A = < u,y > is the observability condition. This is an easy computationand can be left to the reader. Looking for poles and zeros is equivalent to finding a relationlike:

And in this case we have:d d

'dt'd.

'dt'd d. s ) , + 4 ( e 3 + 4 < ' 2 = ( - ( ^ - 4 + 1 4 / '

Therefore, poles are the zeros of the polynomial p(x) = s 4 + 4 i 3 + 4a;2, that is to say,x = 0 and x = 2.On the oth er han d, zeros are the zeros of the polyno mial q(x) =x'2+4x lA and, therefore,are x = - 2 3 \ /2 .

3 . Consider the circuit shown in figure 2.1. The equations are:

X = -1/RiC 00 -R 2L X +1/RiC

1/Ly = ( -l/R i l ) x + l/R lU

u

Using the same method as in the last example, a generator of the module is found: z =~^ cxi + X 2 - So A = < z >. This module is free in all cases except L = R1R2C. In thiscase, (Lgj + R%I)z = 0. That is to say, there are torsion elements in A, and therefore thesystem is not controllable.Let us cons ider t he observab ility condition. It is necessary to check whe ther or not A isequal to < y,u >. The state variables are involved in the following equations:

y \/R\u = ~1/R\X\ + X 2


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2.5. EXAMPLES 21

Vi

walWN

il'1m i

A mK 2Ui

c 2il 1

T712 "2

V2

Figure 2.2: Mechanical system with springs and dampers corresponding to example 4y - l/Ri + ( - ^ - 1/L)U = ^Q X * RlLx2

Th is sys tem ha s a solution if, and only if, L ^ R1R2C. Therefore the system is observableif and only if L ^ RxR 2C.In the case L = RxR2C there exist decoupling zeros. Remember that the input-decouplingzeros are the eigenvalues of the linear mapping induced by j in the torsion submodule T.The generator of this submodule is z and it fulfills the following equation:

1z + RXC z = 0So , -g ^ is the eigenvalue sought.Th e outpu t-de cou pling zeros are the eigenvalues of the linear ma ppin g induced by ^ inS = A generator of this submodule is x\ . In this submodule the equation is

xi = - R\C xxSo, again, the output-decoupling zero is -j^

4. Another example is drawn in figure 2.2 and modelled by the linear system:/

X00fci + k27711

m 2

00

1 2 .m i7712

10 0 \1C 1 + C 2 C 2m\ mi7T12 7712 /

/

x +0 0 \0 01/mi 0o i/m 2 y

17

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22 CHAP TER 2. LINEAR SYSTEMS

Y =It can be shown tha t A = < x\, x2 > This module is free and, for this reason, the system iscontrollable. It is also very easy to check the observability condition A = < ui, u2, j / i , y2 >It is possible to find the following relation between the input and the output variables:If ,d_\ _ mim2(^)4+(miC2+m2Ci+m2C2)(^:) 3+(ciC2+A:2mi+A:im2+fc2m2)(^) 2"\dt' raira2(fc 2ei+fcie2)(^)+fcifc2JrmirU2Then :

rn 2(& )2 + c2() + k2 c2()+k2c2{i) + k2 m ^ ) 2 + (Cl + c2)() + (h + k2)For this reason the zeros are the zeros of the determinant of the last matrix, and the polesare the zeros of p().at

2.6 F i n a l r e m a rk sA formal Laplace tranform and the transfer function matrix are naturally defined in the moduleformalism in [20], where the relationship between left(right) coprime matrix decomposition andcontrollability (observability) is also stu died.Most of the concepts and results in linear control systems have been presented within the framework of the new algebraic formalism introduced by- M. Fliess. In add ition to t he concision,clarity and stylishness of the concepts, this approach is specially appropriate for problems involving tra ck ing of references an d generation of signals. Moreover, this algebraic frameworkenables the classical results to be improved. On the other hand, the concision, clarity andstylishness make it easier to consider some phenomena which have sometimes been ignored inthe control literature and seem difficult to explain in any classical framework, as will be seen inthe next subsections:2 . 6 . 1 T h e m a t c h i n g c o n d i t i o n i n s l i d in g c o n t r o l m o d eLet

X = AX + Bu\y = CXbe a linear single-input system, and assume that we want y = 0 to be achieved as steady state;that is to say, a sliding regime on the sliding surface CX = 0. In classical references ( [62],[16]) the existence of a sliding regime and the description of the ideal sliding dynamics is closelyrelated to the equivalent control (u eq) , which is derived from y = CX = 0. A necessary

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2.6. FINAL REMA RKS 23

u . s - 1s+ 1v . 1s - 1 y ,

Figure 2.3: Not stabilizable examplecondition for obtaining ueq from y = 0 is that the relative degree of y is 1. O therwise ueq cannotbe well defined.In the framework of module theory, the module over K[-jj] spanned by y is considered insteadof the sliding surface equations. In this module an element

0U[y) \u-(y) if U B < 0with u~(y) < ueq(y) < u+(y ) guarantees the achievement of equation 2.3. Finally y = 0 is theassymptotically stable equilibrium solution of equation 2.3. Hence y = 0 is obtained as steadystate and the control objective is attained.2 . 6 . 2 T h e l i n e a r s y s t e m i n t e r c o n n e c t i o n sLet us consider a motivating example. The system:

,(2 ) _ . , =yv y u u (2.4)whose transfer function is s - 1

s2 - 1 s + 1corresponds to the block diagram in figure 2.3, since u = v-\-v,v = y y.Let z = y + y u. T h e n z z = 0. So, z satisfies an unstab le equation . Th is implies th atsystem 2.4 is not stabilizable.The reverse block diagram is figure 2.4.It corresponds to u = w w = y + y. That is to say,

y + y = uIts transfer function is a lso ^ ; bu t this is input-output stable .Finally, let us consider the feedback system in figure 2.5.

(2.5)


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u.

s-1s + l w 1 - 1 ,y

Figure 2.4: Reverse block diagram of 2.3

T{s)

S(s)

V

Figure 2.5: Ill-posed systemIts transfer function is x L . If TS = 1, th en the system is "ill-posed" in the sense of Willems[68].It is difficult to explain these phenom ena in any classic framework. Th ere is no difference betweensystems 2.4 and 2.5 in the transfer function approach. But in the module framework we noticethat system 2.4 is not free torsion because z is a torsion element. On the other hand, system 2.5is free and y is a generator.These kinds of system s, called interconnections, have been examined by M. Fliess and H. Bourls[21] via a standard algebraic tool, coproducts of modules ( [8],[43] ). They confirm Willem'standpoint [64]:" I t is often misleading to distinguish between systems variables".Consider a family of modules {Ma, a E A}. Let E be a given module such that, for any a G A,there exists a morphism:

ha : E > MaDefine the submodu le e of the cartesian product x a y i M a spanned by the elements of the form

(..., 0 , . . . , hai ( e ) , . . . , 0 , . . . , -K 2(e),..., 0,...)where e E E and ai ^ Q.%. The quotient module xaeAM a/E is called the coproduct ( or thefibered sum, or the amalgamated sum ) of the M Q 's(referncies). It is wri t ten Ua.A,EMa.W h e n the modules Ma 's are viewed as linear systems, the above coproduct is called a systeminterconnection. These interconnections are defined without distinguishing between system variables. Some examples are studied below. Let D% be a dynamic with inputs ul = {u\,. .. ,u%m}and outputs yx = {y\,..., y^} for i = 1,2.

1. If m := m1 = m? and p := p 1 = p2, consider the parallel interconnection from figure 2.6.


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2.6. FINAL REMA RKS 25

u1 = u2tD 1

D 2

y1,

v\Figure 2.6: Parallel interconnection

\ D 1 y 1 = u2 D 2 v\Figure 2.7: Series interconnection

Consider the free module [5] = [Su..., Sm] of rank m, an d the two canonical isom orphisms: [u|]e s > uzs

The interconnection is D 1 U yi= t 2 D 2, which is defined by the equations of each moduleplus the equation y 1 = u2.Th e first two cases of the m otivating examples are examples of this typ e of interco nnection .

3 . Let D 3 be a third dynamic with input u 3 = {u \,..., u 3 ^} and output y 3 = {yf,..., y 3 3 } .Consider the feedback interconnection whose block diagram is figure 2.8.The inpu t u3 = v U w is divided into two parts. Set y 3 = it 1, y 1 = it2 , y 2 = w. Therefore,the above block diagram corresponds to the coproduct U y3 := uityi=u 2 iy2 -W(D 1,D 2,D 3).


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Figure 2.8: Feedback interconnectionA frequent phen om ena is the lack of controllability or observability; tha t is to say, interconn ectingcontrollable (or observable) linear systems may give rise to an uncontrollable (or unobservable)one. This cannot be detected by transfer functions. Moreover, when K is a field of constants,the hidden modes corresponding to the lack of controllability (or observability) may exhibitpositive real part s which imply unstability. Let us see the examples state d at th e beg inningof this su bsectio n. As has been said, example 2.4 is not torsion free. In other wo rds, it is notcontrollable. The corresponding input decoupling zero, which is 1, is unstable.In system 2.5, w canno t b e expressed as a linear combination of u, y and a finite number of theirderivatives. So, the system is unobservable.Consider the third example. Write T(s) = | f | j a nd 5 ( s ) = 4 4 , a,b,c,d R[s], abed ^ 0, a,b(resp. c, d) coprime. The system is governed by the equations:

a ( ^ ) ( - w ) b i ) yCJt)y(2.6)(2.7)

There are two possible si tuations:1. If ac + bd 5 0 (i.e. ST ^ - 1 ) , t h e n (b(s)d(s) + a(s)c{s))y(s) = a(s)d{s)u{s). Therefore, y

can be obtained from u; and v can also be obtained from y, and, consequently, from u.2. If ac + bd = 0, then u must satisfy a ( ^ ) u = 0. S o u becomes a torsion element. Th e

remaining variables y, v spa n a free mo dule of rank 1. Here, the lack of cont rollab ilityconcerns the control variable.

Another strange phenomena is the change of rank. Generally speaking the rank is the maximumnumber of independent channels, but may change in some interconnections. Let us consider anexample:The system in figure 2.9 is governed by the equations:

-a{It)v = b^yd(^-)vKdr

= c { t ) y(2.8)(2.9)


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2.6. FINAL REMARKS 27

_ + -V .

V

T(s)

S(s)

V

Figure 2.9: Example illustrating possible changes of rankThere are also two possible situations:

1. If ac + bd / 0 (i.e. ST ^ 1) then (ac + bd)v = 0. This implies that v, and consequentlyy , are torsion. Therefore, the rank is zero.

2. If ac + bd = 0 (i.e. ST = 1) the module is free of rank 1.


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C h a p t e r 3L i n e a r i z a t i o n of n o n l i n e a r s y s t e m sand flatnessThis chapter serves as an introduction to different types of linearizations for nonlinear controlsystems: static feedback linearization, dynamic feedback linearization and linearization by pro-longations, which is a particular case of dynamic feedback linearization. The concept of flatnesswill be also introduced, as well as the concept of flat outputs. The tools and concepts of the twodifferent frameworks u sed throu gh out this thesis will be stated in this chapter. The se frameworksare: differential geometry and differential algebra.

3.1 Different ty p es of l inea riza t ionsD e f i n i t i o n 4 A nonlinear system

mx = f{x) + ^gi(x)ui x G Rni = i

is said to be static feedba ck linearizable if it is possible to find a feedbacku = a{z) + {z)v ueRm ve Rm z G Rn

and a diffeomorphismz = 4>(x)

such that the original sistem is transform ad into a linear controllable systemz = Az + Bv

where A and B are matrices of appropiate size.The next theorem is a characterization of static fedback linearizability in the differential geom-etry framework. A proof can be found in ([34], [31], [51]).

29


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30 CHAPTE R 3. LINEARIZATION OF NONLINEA R SYSTEMS AND FLATNESST h e o r e m 3 Let

mx = f(x) + Y^9i(x)ui1 = 1be a nonlinear system with m inputs. This system is static feedback linearizable if and only ifthe following distributions have constant rank and are involutive:

D 0= i = 1 n_Di= ' ' " ' 'and the rank of -D n- i S nIn the case th at the above s ystem is static feedback linearizable, there exists a change of variablesand a feedback such that the system is written in the following way

Vi = Vi+x Vi = 1 , . . . , n i j kj, j = 1 , . . . , mVkj = Vj j = l,...,m

kj are th e so called Bru nov sky indices [5],[51]. Th e definition of the se indices is as follows: Definepo = dirnDoPi dimDi dimDi-i i > 1

The n ,kj = #{pi >3,i> 0}

A generalization of the st atic feedback is a dynamic feedback transf orm ation.D e f i n i t i o n 5 A nonlinear system

x = f(x,u) xR n ue Rm (3.1)is said to be dynamic feedback linearizable if there exists:

1. A regular dynamic compensatorz = a(x,z,v)u = b(x,z,v) (3.2)

with z Rg and v G Rm. The regularity assum ption implies the invertibility of 3.2 withinput v and output u.2. A diffeomorphism

xp = $(x,z) (3.3)with tp G Rn+q , such that the original system 3.1 with the dynamic com pensator 3.2,after applying 3.3, becomes a constant linear controllable system:


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3.2. FLATNESS AND DIFFERENTIAL ALGEBRA 31This linear system may be written in Brunovsky canonical form ([5],[51]) by means of a staticstate feedback and a linear invertible change of coordinates:

y ki)=Vi V = l , . . . ,mwhere {ki}1l .1 are the Kronecker indices. Therefore, setting

ii - dh i ; ( f c l - 1 ) 7/ ( f c - 1 hy \ylj" iVl ) >y m ) iVm Iit is possible to write y = Tip, where T is an invertible matrix. This can be transforned, usingthe invertibility of the diffeomorphism 3.3, into

A nd from 3.2, u = b('9~1(T~ ly),v). That is to say, x and u can be expressed as real-analyticfunctions of the components of (yi , . . . ,y m ) and their derivatives. The dynamic feedback 3.2 iscalled endogenous if, an d only if, th e converse holds; tha t is to say, if, an d only if, ( y i , . . . , ym)can be expressed as real-analytic functions of x, u and a finite number of their derivatives.Def in i t io n 6 A dy nam ics 3.1 is called (differentially) flat if, and only if, is linearizable viadynamic endogenous feedback. The variables ( y i , . . . , y m ) are called flat or linearizing outputs.Therefore, a flat system is not only linearizable, but is also a system where x and u trajectoriescan be deduced immediately from (yi , . . . , y m ) traje ctorie s. In fact, th is is th e power of flatness.Once the flatness of a system is known, it does not imply that one intends to transform thesystem into a single linear one. W hen a system is flat, it is an ind ication th at the nonlinearstructure of the system is well characterized, and one can exploit that structure by designingcontrol algorythms for motion planning, trajectory generation, and stabilization. Indeed, the flatoutputs are the nonlinear analogue of a basis of the free module for linear controllable systems.Flatness was first introduced by Fliess and coworkers in [13],[14],[19],[22] using the formalism ofdifferential algebra. In differential algebra, a system is viewed as a differential field generated bya set of variables (states and inputs). Recently, flatness has been defined in a more geometriccontext. One approach is to use exterior differential systems, and to regard a nonlinear controlsystem as a Pfaffian system on an appropiate space (see, for instance [49] and referncies therein).A somewhat different geometric point of view is to consider a Lie-Bcklund framework as theunderlying mathematical structure ([23],[24]). In this context, a system is a smooth vector fieldon a smooth manifold, possibly of infinite dimension.

3.2 Fl a tn es s an d d i f ferent ia l a lg eb raFor an introduction to differential algebra see [39],[40],[54].

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32 CHAPTER 3. LINEARIZATION OF NONLINEAR SYSTEMS AND FLATNESSD e f in i t ion 7 Let k be a given differential field. A system is a finitely generated differentialextension D/k. This corresponds to a finite number of quantities which are related by a finitenumber of algebraic differential equations over k. The differential order of the system D/k isthe differential transcendence degree of the extension D/k.Let k the differential field generated by k and a finite set u = ( i t i , . . . , um) of differential/c-indeterminates. Assume ui,...,um differentially fc-algebraically independent; that is to say,diff tr dk /k = m. A dynamics with input u is a finitely generated differentially algebraicextension D/k . Note that the number of independent inputs is equal to the differentialorder of the system D/k as was proven in [63]. An output y = (j/i, . . . ,y p) is a finite set ofdifferential quantities in D.According to theorem 6, ther e exists a finite trascendence basis x = (xi,..., xn) of D/k .Therefore, since ,y D, Xi,yj are k -algebraically dependent on x. That is to say,there exist Ai and Bj, polynomials over k, such that

{ Ai(ii, x,u,u,..., u( r*)) = 0 Vi = 1 , . . . , nBj{yj,x,u,,.. .,u^) = 0 Vj = 1, . . . ,pXi are called generalized states and n is the dimension of the dynamics D/k .As was stated in the former section, linear systems are viewed as finitely generated modules overprincipal ideal rings. The relation between these two approaches (field extensions and modules)is established by what is called Kahler differential ([35]). See appendix A for an introductionto differential algebra and details on the Kahler differential. To a finitely generated differentialfield extension L/K, associate a mapping (the Kahler differential)

dL/K ' L > P L / Kwhere CIL/K is a finitely generated left L[^]-module, such that:

Va GL dL/Kft = i(dL/ Ka)Va,beL dL/K{a + b) = dL/Ka + dL/KbVa, 6 L dL/i({ab) = bdi/^a + adi/^bVceK dL/ Kc = 0

As was seen in the previous section, a module like this corresponds to a linear system. In thiscase, this system is called the tangent or variational system. The inputs of this tangent systemare (d L/KUi,..., dL/Kum). Properties of the extension L/K can be translated into the linearmodule theoretic framework:

A set ijj = (tpi,... ,tpm) is a differential transcendence basis of L/K if, and only if, theirrespective Kahler differentials {di/Ripi,... ALIK^) rnake up a maximal set of L[-^\-linearly independent elements in /A"- I*1 other words, if, and only if, (d L/Ki(ji,... ,dL/Kare a basis of ClL/K. Thus, diff tr dL/K = rk SIL/K-

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3.2. FLATNESS AND DIFFERENTIAL ALGEBRA 33 Th e extension L/K is differentially algebraic if, and only if, the module fi/^ is torsion.

And a set X = (x\,..., xn) is a transcendence basis of L/K (not differential) if, and onlyif, {di/ftX = CIL/K^I, ) dL/Kxn) is a basis of ^LJK a s a -vector space. Th e extension L/K is algebraic if, and only if, ClLfK = {0}.

T he following definition state s precisely wha t dy nam ic endogenous feedback me ans in this framework.D e f i n i t i o n 8 Two systems D \jk andDi/k are said to be equivalent (by endogenous feedback) if,and only if, there exist two algebraic extensions (not differential algebraic) D\/D\ and D2/D2and a differential k-automorphism between D\/k and D-i/k. In other words (identifying thesystems with their respective images in the bigger fields), D\/k and .D2 A are equivalent if, andonly if, any elemen t ofD\ (respectively D and A J A < ^ > are said to be equivalent if, and only if, their correspondingsystems D\/k and D2k are equivalent.P r o p o s i t i o n 2 Two equivalent systems have the same differential order (and, therefore, thesame num ber of inputs). And the same happens to the dynamics.Proof: Let K be the differential field generated by D\ and Di- Since D\ and D2 are equivalent,K/D\ and K/D2 are algebraic extensions. So

diff tr cPDi/k = diff tr dK/k = diff tr d D2/km

Remark: Consider two equivalent dynamics, D\/k and 2k < v >. Let n\ and ri2 bethe dimension of Di/k and D2k < v > respectively. Write the generalized state variablerepresentations of both dynamics:Al(ii,x,u,u,... ,u^r^) = 0 i = 1 , . . . , n iCf{i,z,v,v,...,v^) = 0 = l , . . . , n 2

On the other hand, since any element of D\ is algebraic over D2 and viceversa,' p}(u j > 1 t> l 1 . . . 1 ( i i ) ) = 0 3 = l , . . . ,m

qj{xi,z,v,v,...,vW)) = 0 i = l,...,r , .,pj(Vj,x,u,u,...,uKJ') = 0 j - l , . . . ,mk qf(zi,x,u,,...,u^) = 0 i = l , . . . , r ?

where Pj,p"j, g},g^ are polynomials over k. 3.4 corresponds to two endogenous dynamic feedbacks because they do not use any variable trascendental over D\ and D2.

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34 CHAPTER 3. LINEARIZATION OF NONLINEAR SYSTEMS AND FLATNESS3 .3 L inea r i za t ion by pro lon ga t ionsDef in i t ion 9 Let m

i = f(x) + Y^9i{x)ui = 1be a nonlinear system with m inputs. A prolongation of this system isx = / ( * ) + E i $ i O i ) u ?ul

- 1 =u- Vt = l . ,m

Vi

where u\, which corresponds to u\ , are new state variables Vi = l . . . mAnd the new inputs are V{. j = O . - . & i 1 .Def in i t ion 10 Let

= f(x) + ^2gi(x)uii = l

be a nonlinear system with m inputs. This system is said to be linearizable by prolongations ifthere exists a prolongation of the original system which is static feedback linearizable.In fact, a system which is linearizable by prolongations is dynam ic feedback linearizable. T ha t isto say, a linearization by prolongations is a particu lar case of dynamic feedback linearization. Letus see the relationship between these two types of linearizations. Consider a dynamic feedbackcompensator, affine respect to the inputs:

z = a(x,z) + a1(x,z)vu = b(x,z) + b l(x,z)v (3.5)with z G Rq and v Rm. A dynamic feedback compensator is a prolongation if, and only if,

u =zki+l

Zi =Zi+i if i kj, j = l , . . . ,mVj if i = kj, j = 1 , . . . ,m

where 1 < i < q.

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3.3. LINEARIZATION BY PROLON GATIONS 35That is to say,

/ * i

b(x,z) = Zk!+li+Er=Tlfci /

b\x,z) = 01V ' ' 1 0 if = kj, j = 1, . . . , mflir, , ^ - J i f * ^ A i = ! . mm

The following lemma will be used in some proofs related to linearization by prolongations inchapters 4, 5 and 6.L e m m a 5 Let

7 j - , i G Ia family of coordinate vector fields in R ;

D 1 = < a 1 , . . . l a j ; { } 0 ^ 1 >D2 = < {-5^}kn+l+l >

whered_ _d_dy i ' " * * ' dy nai,..., ctj e< -^-,..., -57- >and not depending on the variables

Vn+l+l,- ,Vn+sThen D\ @ D% is involutive if, and only if, Di is involutive.Proof: It is straightforward.


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36 CHAPTER 3. LINEARIZATION OF NONLINEAR SYSTEMS AND FLATNESS


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C h a p t e r 4L in ea r i z a t io n u s in g d i f fe r en ti a la l g e b r a4 . 1 I n t r o d u c t i o nThis chapter deals with the problem of linearization of nonlinear control systems; that is tosay, with the problem of flatness. Using the methods of differential algebra, we explain theconditions which in this framework must be satisfied in order for a nonlinear control system tobe l inearizable. First of al l , the tangen t system is com puted using the Kahler differential. Th en ,for a single-input system, we give a new proof of the fact that a single-input system is staticfeedback lineariza ble if, a nd only if, i t is dyn am ic feedback linearizable. W e also tackle th e sta ticfeedback l inearizabil i ty problem for a mult i - input system, thereby ob taining the condit ions th ata system must fulfill in order to be transformed into a linear one in this context. The problem ofdynamic feedback linearizability is solved by trying to guess m 1 flat outputs and computingthe last one. Th e quotient of modules appears to be a cornerstone in this procedu re. Final ly,without computing the tangent system, a procedure based also on guessing m 1 f lat outputs isbe designed. The main tool here will be the intermediate differential field extensions. A helpfulsoftware packag e in Ma ple V is created in order to simplify the co mp utation s require d. T helist ing of this program can be found in appendix B.

4 .2 S ing le - inpu t sys t em sConsider the single-input system:

x = f(x,u) (4.1)where the s ta te x = (x\,... ,xn) G Rn and the control u R. Assume tha t 4.1 is control lable.Thanks to the proper ty

d(a) = (da) Va G L37


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38 CHAPTER 4. LINEARIZATION USING DIFFERENTIAL ALGEBRAwhere d is the Kahler differential and L is the field extension correspond ing to 4.1 , the tang entsystem is:

dx = -^-dx + -^-du (4.2)Therefore, the basis of the corresponding module Q contains just one element. Let w be such anelement. In this context, a characterization of flatness for a single-input system is shown, andconsequently the well known equivalence between dynamic and static feedback linearization forsingle input systems can be deduced (see [7], [2], [49] for other proofs of the same result):Propo s i t i o n 3 : System 4-1 is static feedback linearizable if and on ly if the mod ule f2 characterizing 4.2 is generated by an integrable one form w.Proof: The necessity is obvious.Sufficiency: Since w is assumed to be a generator:

dxi = q l;{-)w Vi = l , . . . , n (4.3)where

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4.2. SINGLE-INPUT SYSTEMS 39

WW

because, if it is zero, the above equation becomes

H-(s).(*which implies tha t w is a torsion element, in contradiction with the controllability of the system.Now, since ( ) , * " it is possible to isolate du :

So, Jdu=p li+1 ()w (4.5)atwhere plj+1(-^) is a polyno mial in the indeterm inate ^ of degree lj + 1 > n.Note that it is no possible that

du=ps(4:)w (4.6)dtwith degree s < n, bec aus e if 4.6 hold s, the n, equalling 4.6 and 4.5:p lj+l(-r)w = du =p s()wy Kdt ' F Kdt '

which implies that w i s a torsion element, in contradict ion with the controllabil i ty hypothe sis.Summarizing, the relative degree of w with respect to du is lj + 1 > n. But, it is well-knownth at the relative degree is always smaller tha n or equal to the dimension of the st ate . Therefo re,the relative degree of w with respect to du is n.Now, using the integrability condition of w, a variable y (the flat output) such that dy = w isobtained. And since the Kahler differential commutes with the time derivative, this variable y,the flat output, fulfills the relative degree condition with respect to u . Corol lary 1 : Linearization by static and dynamic feedback are equivalent for single-inputsystems.Proof: As already stated in chapter 2, flatness (or linearization by dynamic feedback) is equivalent to the existence of an integrable basis of the tangent system. Prom the proposition thislast fact is equivalent to linearization by static feedback. Hence, dyna mic and static feedbacklinearization are equivalent for single-input systems.

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40 CHAPTER 4. LINEARIZATION USING DIFFERENTIAL ALGEBRANow, imposing the former relative degree condition, w is computed as a solution of an homoge-nous linear system. Since w lies in fi the expression of w must be:

w = aidx\ + . . . + andx n := a dXAnd an expression for w is obtained:

vj = dx + a (g-dx + g{du\= ( + a) dx + a du

Therefore, the condition of the relative degree implies:- I ; - 0 < 4 - 7 Deriving 4.7, it is possible to compute another condition, useful in the following steps:

du -m -The value of w is calculated:

+ { + a%)ILduThe condition of the relative degree leads to:

f + a^V = 0V dx) duwhich thanks to equation 4.8, becomes:

a.(l_I)d l = Q\dx dt J duHere it is also possible to obtain another useful expression:

"5!(59-- mImposing once again the relative degree condition, now to to'3) :( (2) o-dj_ ICI (2L\2\ QL-oI dx dtdx \dxj I du

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4.3. STATIC FEEDBAC K LINEARIZATION OF MULTI-INPUT SYSTEMS 41Using 4.8 an d 4.9, this last equ ation becomes:

\dx dt J duI terat ing this process, a system of equations is obtained: ( - ) ' - - -This is an homogeneous system with n 1 equations and n un knowns ( a\,... ,a n ). So, thesolut ion w will depend upon a certain function A. Now, a variable y such that dy w must beobtained. The role of A is the role of an integrant factor.

4 .3 S t a t i c f eedback l i nea r i za t ion of m ul t i - i np ut sys t e m sOnce again the intent ion is to compute the basis of the module corresponding to the tangentsystem. As was state d in chapter 2, this basis contains the same num ber of elements as th enumber of the inputs . Let m be this number. And let w\,..., w m be the elements of this basis.And again, as in the single-input case, the conditions that must be imposed to find the basis arethe relat ive degree condit ions. Th e fol lowing lemm as wil l t rans late th e relat ive degree cond it ionsof a nonlinear system to i ts tangent system.L e m m a 6 For a ll f, g vector fields, and for any w differential one form, the next e quality issatisfied:

L f = < LfW, g> + (4.10)L e m m a 7 The Lie derivative with respect to a vector field f comm utes with the exterior derivative:

L f(dh) = dL f{h ) (4.11)The proof of these two lemmas can be found in any elementary text on differential geometry.L e m m a 8 For any C function y and any vector fields f and h, it follows the equality:

< dLrfy,h >= ( - l ) f c ( W * < d ^ a d ) h > ( 4 1 2 )Proof: I t wil l be proven by induction. For r = 1, and than ks to 4.11

< dLfy, h > = < Lfdy, h >App lying now 4.10, we have

< Lfdy, h>=L f -< dy, [/, h] >

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42 CHAPTE R 4. LINEARIZATION USING DIFFERENTIAL ALGEBRAwhich is the desired equality for r = 1.Assuming the trueness of the statement for the case r, the case r + 1 will be proven. App lying4.11 and 4.10:

< dL rf+1 y, h >=< LfdL rfy, h>=L f < dL rfy, h>-< dL rfy, [/, h] >Using the induction hypothesis, the latter expression becomes

h E - W W * < dV>) - D - W W * < dy,adkfadfh >)which, in turn, is equal to

( ( - ! ) * ( ) L ) + 1 ~ k < *V> a d ) h A - ( V i ) * ( ) LT" < dV, if + 1 >)Finally, the equality hull H T

L ' W \ fi/ 1 / \ A, ,applied to the former expression, leads to'kj\-l)k^liyrf+1-k^

This is the desired expression for the case r + 1.P r o p o s i t i o n 4 When a system

x = f{x) + Yu9i{x)uii = lis linearizable by static feedback, then

yi = L rfyi V r < k{ - 2where yi and its derivatives up to order fc 1 are the coordinates of the change of variables thatlinearizes the nonlinear system; and ki are the Kronecker indices.Proof: Again it will be proven by induction. For r = 1

j nyj =< dyj,f{x) + Yjgi{x)ui > = < dy jyf >

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4.3. STATIC FEEDBA CK LINEARIZATION OF MULTI-INPUT SYSTEMS 43since the static feedback linearization conditions imply

< dyj, cufygi > = 0 Vr < kj - 2 (4.13)Assuming that the statement is true up to r , the case r + 1 will be studied. Using the inductionhypothesis we have (F+i n

d~F^Vi = J t ^ ^ =< dLTfVJ>f(x) + J29i(x)ui >Ap plying now the equation 4.12, we get:

( - l ) f c ( T W * < dyj,adkf(f(x) + J29i(x)u i) >which tha nk s to 4.13 can be writ ten

L rf+2(-l)fcf l J L T k u i < d W ' a d )S i >fc=i W =iAgain, because of 4.13,

(-i)*K W * i < foi*/* >= ojfc=i W = i

Therefore,^ r r + 1

Corol lary 2 TTie Kahler differential of yj does noi depend on the inputs of the tangent systemdui i = 1 , . . . , mProof: In the previous proposition it has been proven that ^FTTVJ does not depend on the inputvariables it Vi = l , . . . , m . Th e comm utativity between the t ime derivative and the Kahlerdifferential was commented on in chapter 2. Putting them all together, the statement of thecorollary is derived.So, the relative degree of Wj = dyj with respect to du must be fc, where ki are the Kroneckerindices ([51]).Let x = f(x,u) xeRn ue Rma multi-input system. Its Kahler differential is

dx = ^J-dx + ^-du (4.14)ox ou

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44 CHAPTER 4. LINEARIZATION USING DIFFERENTIAL ALGEBRAThe basis can be wri t ten

Wi a\dx\ + ... + azndx n = a1 dx Vi = 1 , . . . , mLeading the relative degree condition to

' ( i - l ' T s ; - 0 w - " ." .* - 2 = !,...,The solution of this system will be a basis of fi . Note that, if the Kronecker indices are suchtha t k\ > k-2 > . . . , then w\ depends upon, at least , n (k \ \)m parameters ; IV 2 depends a tleast upon n fa l)m param eters; and so on. These param eters act as integrant factors inorder to find y G L such that dyi = iu.Note that even though the original system was not affine in the inputs, this procedure can stillbe appliedExample: Let us consider the following system:

1 = 2 + ^ 1 ^ 22 = Z3U1U2 + X2U23 = U2Z3

I ts tangent system is:dx\ = uzdxi + dx2 + x\du2

< dx 2 = U2dx2 + 2 xzU\U2dxz + x\v,2du\ + {x2 + x1u\)dv,2dx z = U2dx$ + xzdu2

The Kronecker indices are k\ = 2 y &2 = 1.DenotingWi = a\dx\ + a\dx2 + a\dx^ = 1,2

Applying the conditions explained above, the next equations for w\ are obtained:a\ = 0 and a\x\ + 0,3X3 = 0

and no condit ions for W2 - Thereforeu>i = \{xzdx\ x\dxz) A G L

and IU2 can b e choosen freely, ta king into a ccount th at W2 has to be differentially ind ep en de ntwit h respe ct to 101. As A is an inte gran t factor for w\, it can be chosen in such a way that w\is an exact one form. A possibility is to choose

2:1X3


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46 CHAPTE R 4. LINEARIZATION USING DIFFERENTIAL ALGEBRA

"i = H^)dyi + + q^(jt)d ymwhere R^ ( ^ ) , . . . , R l (-^) are vectors with coefficients in K[(-^)] and h,...,lm are the maxi-mu m degrees of the indete rm inate in each vector, f / 1 ^) + . . . + ?,>"( g) are polynomials inthe indeterminate (J) and fin,..., h{m are the degrees of the respective polynomials.Now

Jb . . ._ J kJi i f P j i ^ O]l ' 1 rel d(dyj,dui) if pji = 0rp := max{{Zj + kj p,V j = 1 . . . m } , {/i,- + kj p,V j = 1 . . . m , Vi = 1 . . . m } }

rij := max{Zj, /ij, V = 1 , . . . , m }It must be noticed that Z := oo (respectively fty = oo) if Rj = 0 (respectively gy = 0).Corol lary 3 7ie system is dynamic feedback linearizable by prolongations if and only if thesets V and W have the same number of variables, where W = :

{dxi,..., dxn, dux,..., duY1',..., du m, .., du^}an d

V = {dyu...,dy^\...,dy m,...,dydy^\...,dym,...,dy^And conversely, d y i , . . . , d y i n i ) , . . . , d y m i . . . J d y ^are linear functions of

dx\,..., dxn,dui,..., dui,..., dum,..., du^m'In o ther w ord s, the re exists a linear change of variables between the variables in V and W . T h i schange of variables is, in fact, thanks to the integrability conditon, the jacobian of the changeof variables between

X\, . . . ,Xn, U l , . . . , U , U m , . . . , U m

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4.5. SOFTWAR E 47and

/1 > ) y i ) i ym > / mSo, as the change of variables exists, the system

x = f(x,u)faW = u j+1) V* = l , . . . ,m V ^ O , . . . , ^is static feedback equivalent to

J tVi)=Vi+l ) Vi = l , . . . ,m Vi = 0 , . . . , n < - 1

x = f{x,u)In other words, the original systemis linearizable by prolongations.4.5 SoftwareIn order to simplify the calculations needed in the method presented above, a software packagehas been developed. It consists in many Maple V functions that allow us to use a computer toperform the basic operations -such as the computation of the tangent system- and those that areharder. For example, given a single input system, a basis of the module fi can be determinedand quotients in this module with arbitrary expressions can also be made. Given a two-inputdyna mic feedback linearizable system, it can be reduced to a single input one thro ugh a quo tient,and, in this single input system, a basis can be computed.The integrability of the basis is automatically tested, and, if it holds, the integrals will be theflat outputs.4 . 6 E xa m pl e s

1. This example has been borrowed from [45].Let:

1 = 1l2 = 112 3 = UiU2be a nonlinear system.

The tangent system associated to it is a quotient "[^]-module A defined by the generators{dx\,dx2, dx3,dui, duz]


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48 CHAP TER 4. LINEARIZATION USING DIFFEREN TIAL ALGEBRAand the relat ions: dx \ = du \

dx2 dii2dx = U2du\ + u\du2

dx \ can be guessed as one of the two generators of the free m odule. Let us considerthe quotient module i = . In the state space representation, Q,\ is given by theequations:dx2 = dv,2dx$ = U\du2

Clearly, a basis of 2i is given by UJ, = u\dx2 dx%.Going back to 102 = U\dx2 dxz + p(^)dxi; where p(jjj) K[j^\. In order for W2 to beif-integrable, an appropriate choice for p(^) is

d dp{di ) = X2d-t

Hence, an "-integrable basis of 3 isCJI = dx iCJ2 = U\dx2 dx 3 + X2du\

and the flat outputs are:Vi = xiy2 = U1X2 - 3

The relationship between the state variables of the tangent system and the basis of thetangent module is:

dx \ = u) \dX2 =

(2) , X2W\ + (JL>2

Ui (2 )X2U1U1 UiX2W\ U1W2 + UiU)2dx z = :du\ = (idu 2 = involves thi rd derivatives of wiand second derivatives of UJ2


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4.6. EXAMPLESTherefore

n = m a x { l , l , 0 , - o o , 2 , 2 } = 27 -2 = max{00, 1, 00 , 00, oo,0}m = max{ 2,1 ,3} = 3ri2 = m ax {l , oo,2} = 2

Summarizing, the nonlinear system defined by

xi = t i n 2 = V2

i l = 1 2U2 = 1

i s s t a t i c f eedback equiva len t to the sys tem2/11 = 2/122/12 = 2/132/13 = 2/142 / 2 1 = 2/222 / 2 2 = 2/23

which is linear.2. This example has been borrowed from [53].

X2X1 x\2 = 2:3X\X$ = X4

This system can be written down in the following way:x\X23

===

x a + x i u i1 2I t lI I


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CHAPTER 4. LINEARIZATION USING DIFFERENTIAL ALGEBRAIts tangent system is:

dx = 0S 3 + zm i 1 \

00

X 200

(+

1 0X 21 0 | du

i i

Making quotient by dx\ Adu2 ( that is to say, x\ Xx^ in the original system ) we get:/ U l _x 3+ ii _L \

CX =V

0 00

1 200

da;+

Applying the algorithm for single-input systems, the basis u>2 is obtained:/ x2{u2\-xi) - u2X-xi -W2=ai( 2\^rdxi+~2^r+dxz)

where ai is a function depending upon the variables x' s an d u's. Therefore:W2=W 2+p() (dxi - \du2)atBut this one form is not integrable. So, x\ \x in the original system cannot be a flatoutput . If A is zero ( that is to say, guessing x\ as a flat output ), the quotient is:

dx2dxz

jca+nm X2X\X2 X\X20 0

+ ) 2dx2dxz

Clearly, dx2 is a basis of this module. Therefore, the flat outputs are x\ and X2-Comparing this method with the method used in [53], note that our algorithm reduces to:

(a) Making a quotient of modules.


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4.6. EXAMPLES 51(b) Solving an homogeneous linear system.(c) Checking the integrability condition of just one form.

Therefore, it seems simpler than the procedure appearing in [53] .3. The following example is related to a vertical take off and landing aircraft (VTOL). Thisaircraft was assumed to b e not linearizable by prolongations since [23], bu t it was knownto be flat . In our framework we prove not only that is flat, but that it is also linearizableby prolongations. Here are the equations of the system:

x = u\ sin 6 uze cos 6y= uicos6+ U2sin0 10= u2

Reducing the system to order 1 in order to be able to apply our algorithm:x\ = X22 = uisinrE U2ecosrc53 = X44 = 1+ icos:E5+ uiesmxs6 = U2

wherex\ = x X2 x X3 = y X4 = y x$ = 6 XQ 6

As in the preceding examples, its Kahler differential is computed:dx \ = dx2dx2 = (u i cos s + U2s'mx5)dxs+ sinxsdui ecosa;5du2dxs = dxidxi= ( sin^ + U2Cosx5)dx5+ COSX5CU1+ esinx5


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52 CHAPTER 4. LINEARIZATION USING DIFFERENTIAL ALGEBRAA basis of this quotient module is

_ _ = ?^^^lXdxi + Xdx2+

W h e nA = XQ COS X5

^ ) is the appropiate polynomial,2x1 sin 15 +U2 cos i s ,/ 2 l 2 s i n i 5 + U 2 C O S l 5 , \ , .y 2 = X\ y-* 2 ) + X2X6 c o s Z5 +

2igcosi5- U2sin a:5 , _ .Xs B g 1 X4XS S 1 I 1 5Once the flat outputs have been obtained, we are able to decide whether or not this systemis linearizable by prolongations. For this purpose the parameters previously defined arecomputed : i = 3 h = 5

h\\ = 4 h\2 = 6 /121 = 00 /122 = 2kn = - 4 12 = 0 &21 = - 0 0 fc22 = - 2

Therefore,an dThat is to say,While

ri =0 T2 = 4n i = 4 ri2 = 6

n + 7-i+r2 = 6 + 0 + 4 = 101 + ri2 = 10

Therefore, following the corollary, the VTOL is linearizable by prolongations.4 .7 A n ana logou s p ro ce du re us ing fie ld ex tens io nsThe procedure explained in section 3.3 admits a nice counterpart using only field extensions.Let us consider a nonlinear control system

x = f(x,u) xER n ueR m

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4.7. AN ANALOGOU S PROCEDU RE USING FIELD EXTENSIONS 53As was stated in chapter 2, this is a differential field extension L/R, where L is the minimumfield containing the variables x, u and where the equations of the system are satisfied. If m 1flat outputs yi,... ,ym are guessed, let us consider the intermediate differential field extensionL/R < J / 1 , . . . , Vm-i > /R - Then, the extension L/R < y i , . . . , y m -i > is a single-input system.In this single-input system, the relative degree condition can be applied in order to find its flatoutput , which wil l be the remaining flat output yn of L/R. This procedure has a disadvantagewith respect to the one explained in section 3.3. The procedure in section 3.3 uses quotientof modules, and because of this, some equations and m 1 inputs are eliminated. In the fieldextension proce dure , neither equations nor variables can be eliminated. An othe r difficulty isthat there is no software package to apply the procedure, and neither is there a sistemtic wayto obtain the last f lat output .Example: Let us consider again the VTOL. Let us also recall that the system equations can bewritten as follows:

Xl =x2 = 3 =x\ = 5 = 6 =

X2X- 1 +X$

Ui s i n 5U\ COS 5 +

U 2E COS 5t o c s i n 2 5

UlIf yi = X5 is guessed as one of the flat outputs, the extension R < yi,V2 > /R < Vi > isrepresented by the following system

Xl = 2 = 3 =X4 =

X2ui s i n X5 U2E c o s X5X4 I + U 1 C O SX 5 + U 2 s i n a : 5

which is a single-input system. Applying the usual static feedback linearizability conditions tothis single-input system, the following distributions have to be involutive:

> , =

is trivially involutive.D 2=\9u92]] +Vn+lVn-H\gi,\gi,Sn]]


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58 CHAPTER 5. LINEARIZATION BY PROLON GATIONS OF 2-INPUT SYSTEMSL e m m a 11

x _ d d d d ,ni+i ^ o ' i p, ' p, i ' a i Q-Q-fC J / n + n i C y n + m n 2 i uy n+ni+n2 Oyn+ni+i /

ftere ni2 o1=1 + 9lVn+\ + 2_s Vn+j+i:j=1 dyn+ j

Proof: The statement is clear for all i 1^2)) then the system is linearizable by prolongation of u\ n\ n2 times.Proof: It is the result of applying the former proposition n 2 times. T h u s , in the following, prolo ngatio ns by derivatives of ju st one inpu t will be tak en in to consideration for two input systems. The same proof can be done for a system with m inputs, whereonly m 1 have to be prolonged.


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5.2. MAIN RESULT 595.2 M a i n r e s u l tThe following theorem, which establishes a necessary and sufficient condition for the existence ofprolongations, also provides a finite algorithm to decide whether or not a system is linearizableby prolongations.Th e o r em 4 The system

x = f{x) + gi{x)u1+g2(x)u2 x E Rnis linearizable by prolongations if and only if one of the following systems is static feedbacklinearizable:

( x = f(x)+gi(x)y n+ i+ g2(x)w 2Vn+j - Vn+j+i Vj = 1 , . . . , k - 1Vn+k = V)ior _ x = f(x)+g2(x)yn+ i+ gi(x)wiSfc : yn+j = Vn+j+l Vj = 1 , . . . ,fc- 1Vn+k = W2where k = 1 , . . . , 2n 3 and

yn+j = u^~1) j = l,...,k(or, respectively) yn+j u2 j = 1, , kar e the new state variables and

w\ = u\ w2 = u2 respectively w\ = u\ w2 = U2 )ar e the new inputs.Proof: It will be proven that the static feedback linearizability conditions for the system:

2n-3 : * x = f{x)+g1{x)yn+ i+ g2{x)w 2Vn+j = Vn+j+l Vj = 1 , . . . , 2n - 42/3n-3 = Wiand the static feedback linearizability conditions for the system:

x = f(x)+gi(x)y n+ i+ g2{x)w 2s i : S Vn+j = Vn+j+i Y J = 1 , . . . , I -1Vn+i = m

with I > 2n 3 are equivalent. So, this being proven, if a system is linearizable by prolongationsadding I derivatives of u\ (I > In 3), then it will also be linearizable by prolongations adding

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60 CHAPTER 5. LINEARIZATION BY PROLONGATIONS OF 2-INPUT SYSTEMSonly 2n 3 derivatives of i t i , and, obviously, the same fact occurs with U2- Therefore, a finitealgorithm for checking the linearization by prolongations will be to check the static feedbacklinearizability conditions for S and E, VA: = 1 , . . . , 2n 3. As the same proof is valid for bo thtypes ofprolongations (resulting from adding derivatives of u\ or U2), it will be proven just once,in this case for prolongations of u i . The details for the prolongations by derivation ofU2 can berewri t ten in the following proof by changing u\ by U2, and viceversa.So, let 2 n - 4 p.f2n~3{x,y) = f(x)+gi(x)yn+1 + ] Vn+j+i^

j = 1 OVn+jb e the drift associated with the system S2n-3> and let

9 n-3(x,y) = -^ g22n-2(x,y) = 92{x)be the its control fields.I2n-3 1S static feedback linearizable if and only if the following distributions are involutive andconstant rank:

DT~Z = Denote

T]k = ad kpn-392 VA; > 0The following lemmas clarify what {% k > 0} are.L emm a 13 rjk e S =< >Proof: It is proven by induction. For i = 1,

2 n - 4 m = [f + 9\Vn+i + X) vn+j+i-g--,92) = [f,92 + yn+i[gi,g2] e S

Assuming T ] E S, i 7 n - l >

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5.2. MAIN RESULT 61L e m m a 1 4 rjk depends only on the variables x,yn+i,-,Vn+k-Proof: As in the previous lemma,

Vi = [f,92]+Vn+ i\gi,92]On the other hand, Vk+i = [f2n~3,Vk{x,y n+i,---,yn+k]where the induction hypothesis has been applied. Therefore, i t is clear that

p d dVk+l = [f + Vn+l9l + 2 _ , Vn+j+l-g- -, r]k] + [Vn+k+l-Q- ,Vk]which depends only on the variables x,yn+i, j Vn+k+i-Let us now compute ad 3pn_3 g .L e m m a 1 5 1.

1 Oyzn-Z Oyzn-z-j

Proof:1. For j = 1,

ad2-%- = -gi1 Oyzn-3

d . d d . dadpn--z = [yzn-3-z 1 J pn-3-z = iy3n-3 ) "5 J - 1 dyzn-z oy3n -2 dy3n -3 oyzn-2Assuming the equali ty is t rue up to j ,

1 OyZn-3 ! Oyzn-Z Oy3n-3-jwhere the induction hypothesis has been applied. This Lie bracket is equal to

2n4 o a o[f+9iyn+i+ ^+i+i5r-(-1) ir l = (-1) i+ 1sr-2 1T oyn+ j dyzn-3-j d y 3 n - 3 - j - i2. Prom the former proof we have

a < f c 4 d_ a2 n - 3 "2 n y 3 n -3 dy n+ iTherefore,

7 y 3-3 oyn+i

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62 CHAPTER 5. LINEARIZATION BY PROLONGATIONS OF 2-INPUT SYSTEMS

It has been pointed out that 77n G< 770,... , 77n_i > But n may not be is not the least integersatisfying such property. So, let us definer = min{k\r]k e< Vo, ,Vk-i >}

There are two possibilities for these distributions:1. r = n2. r < n1. If r = n, the distribution D^[3 is equal to

d d< ? o , , % - i ! ^ > i n >Oy3n-3 OJ/2n-2

Thus , this distribution being involutive, D].n~3 are involutive for all k > n. The reasonfor this fact is t ha t

r^2n-3 d 9oyzn-z oyzn-z-k

and the Lie brackets[ , Tfr] G 5 =< 770,. -., 77_i >

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